Q & A STPM 2022 Physics

Physics Semester 3 STPM Chapter 21 Sound Waves
1 T
(ii) Fundamental frequency, f 0 =
2L μ

= 1 4.80 Hz
–3
2(2.00) (4.80 × 10 /2.00)
Penerbitan Pelangi Sdn Bhd. All Rights Reserved.
= 35.4 Hz

Three lowest frequencies: f 0 , 2 f 0 , 3 f 0
= 35.4 Hz, 70.8 Hz, 106 Hz




21.3 Intensity Level of Sound


Section A Multiple-choice Questions

Question 1

A sound wave has an intensity of 4.00 μW m . What is the intensity level of
–2
sound?
(Threshold of hearing: 10 W m )
–12
–2
A 25 dB C 66 dB
B 40 dB D 76 dB

Answer: C
I
Intensity level of sound, b = 10 log   dB
I 0
Common Error = 10 log  4.0 × 10 W m –2 dB
–6
–2 
–12
Distinguish the difference 10 W m
between intensity and = 66.0 dB
intensity level.

Question 2

The power output from a sound source emitting a single frequency is doubled.
What is the increase in decibel level?
A 0.50 dB Semester
B 2.0 dB
C 3.0 dB 3
D 4.0 dB



411




021 STPM Q&A Physics T3.indd 411 17/02/2022 10:24 AM

Physics Semester 3 STPM Chapter 21 Sound Waves
Answer: C
I
b = 10 log  
I 0
When power output is doubled, intensity I is doubled.
2I
Penerbitan Pelangi Sdn Bhd. All Rights Reserved.
b’ = 10 log  
I 0
I
= 10 log   + 10 log 2 = b + 3.0 dB
I 0
Question 3

The intensity level of sound at a distance of 3.0 m from a source is 60 dB. What
is the sound intensity level at a distance of 5.0 m from the source?
A 14 dB C 36 dB
B 22 dB D 56 dB


Answer: D
1 I
Intensity, I ∝ , intensity level, b = 40 dB = 10 log  
r 2 I 0
When r = 5.0 m, I’ = 3.0 2 I = 0.36I
5.0 2

b' = 10 log  0.36I 
I 0
I
= 10 log   + 10 log 0.36
I 0
= 56 dB


Section B Structured Questions
Question 4

The power output of a radio is 12.0 W. The radio emits sound equally in all
directions.The threshold of pain is when intensity = 1.00 W m . At what
–2
maximum distance from the radio would the sound be painful to the ear?


Answer
P
Intensity, I = 4πr 2 = 1.00 W m –2
12.0
Semester
3 r = 4π(1.00) m
= 0.98 m


412




021 STPM Q&A Physics T3.indd 412 17/02/2022 10:24 AM

Physics Semester 3 STPM Chapter 21 Sound Waves
Question 5

The intensity level of sound at a distance of 3.00 m from a source is 120 dB. At
what distance is the sound intensity level (a) 100 dB and (b) 10.0 dB?

Penerbitan Pelangi Sdn Bhd. All Rights Reserved.
Answer
I
(a) r = 3.00 m, b = 120 dB = 10 log  
I 0
I = (10 )(10 –12 W m ) = 1.0 W m –2
–2
12
When b = 100 dB
–2
10
I = (10 )(10 –12 W m )
1
= 10 W m ∝
–2
–2
r 2
1
1.0 W m –2 ∝ 3.00 2
r = 1.0 (3.00 m) = 30.0 m
10 –2
(b) When b = 10 dB, I = (10 )(10 W m ) ∝ r I 2
1
–12
–2

I
–11
–2
10 W m ∝
r 2
1
1.0 W m –2 ∝ 3.00 2
1.0
r = (3.00 m) = 94.9 km
10 –11

Section C Essay Question

Question 6

(a) Distinguish between intensity of sound and intensity level of sound.
(b) Sound is emitted from a small speaker uniformly in all directions. Points P
and Q are at distances of r and 2r from the speaker.
Compare Semester
(i) the intensity of sound at P and Q,
(ii) the sound intensity level at P and Q.
3




413




021 STPM Q&A Physics T3.indd 413 17/02/2022 10:24 AM

Physics Semester 3 STPM Chapter 21 Sound Waves

(c) Two small speakers emit sound waves of different frequencies equally in all
directions. Speaker A has an output of 2.5 mW, and speaker B has an output
of 4.50 mW. Determine the sound level at point C in the figure.
C
Penerbitan Pelangi Sdn Bhd. All Rights Reserved.

4.00 m
A B






3.00 m 2.00 m
(i) when only speaker A emits sound,
(ii) when only speaker B emits sound, and
(iii) when both speakers emit sound.


Answer
(a) Intensity, I is the energy per second incident normally on a unit surface
I
area. Sound intensity level in dB = 10 log   where I is the intensity, I 0
I 0
is the threshold of hearing 10 W m .
–12
–2
1
(b) (i) At P, I P ∝
r 2
1
At Q, I Q ∝
(2r) 2
Hence I P = 4I Q
I Q
(ii) Intensity level at Q, b Q = 10 log  
I 0
Intensity level at P, b P = 10 log  (4I Q ) 
I 0
I Q
= 10 log   + 10 log 4
I 0
b P – b Q = 0.60 dB
Intensity level at P is 0.60 dB greater than intensity level at Q.

Semester
3




414




021 STPM Q&A Physics T3.indd 414 17/02/2022 10:24 AM

Physics Semester 3 STPM Chapter 21 Sound Waves
(c) (i) AC = 5.00 m
P 2.50 × 10 –3
–2
I A = = W m = 7.96 μW m –2
4π(AC) 2 4π(5.00) 2
–12 
b A = 10 log  7.96 × 10 –6 = 69.0 dB
10
Penerbitan Pelangi Sdn Bhd. All Rights Reserved.
(ii) BC = 2.00 + 4.00 = 18.00 m 2
2
2
2
P 4.50 × 10 –3
–2
I B = = W m = 19.9 μW m –2
4π(BC) 2 4π(18.00)
–12 
b B = 10 log  19.9 × 10 –6 = 73.0 dB
10
(iii) b A+B = 10 log  (7.96 + 19.9) × 10 –6  = 74.4 dB
–12
10
21.4 Beat


Section A Multiple-choice Questions

Question 1
Two tuning forks are sounded at the same time, 30 beats are heard in 5 seconds.
If one of the tuning forks has a frequency of 252 Hz, what are the possible
frequencies of the other tuning fork?
A 246 Hz or 258 Hz C 249 Hz or 255 Hz
B 247 Hz or 257 Hz D 237 Hz or 267 Hz


Answer: A
Beat frequency = 30 Hz = 6 Hz
5
f 2 – 252 = 6 Hz, or 252 – f 2 = 6 Hz
f 2 = 258 Hz or f 2 = 246 Hz


Question 2
A tuning fork of frequency 262 Hz is sounded along with a guitar string. Four
beats are heard every second. Next, a bit of wax is melted onto the tine of the Semester
tuning fork. The tuning fork now produces five beats per second with the same
guitar string. What is the frequency of the string?
A 257 Hz C 262 Hz 3
B 258 Hz D 266 Hz



415




021 STPM Q&A Physics T3.indd 415 17/02/2022 10:24 AM

Physics Semester 3 STPM Chapter 21 Sound Waves
Answer: D
Beat frequency, 4 Hz = (262 – f) Hz, or 4 Hz = (f – 262) Hz
With wax on its tine, frequency of tuning fork is less than 262 Hz.
Beat frequency increases to 5 Hz, hence the correct equation is
4 Hz = (f – 262) Hz
Penerbitan Pelangi Sdn Bhd. All Rights Reserved.
Frequency of string, f = (262 + 4) Hz = 266 Hz



Section B Structured Questions

Question 3

Two identical strings of length 0.700 m have the same fundamental frequency of
440 Hz. The tension in one of the strings is then increased by 2.0%. The strings
are now struck, what is the beat frequency between the fundamentals of the
two strings?


Answer
T
The speed of wave in a stretched string, v = where T = tension, μ = mass m –1
μ
v 1 T
Fundamental frequency, f = =
2L 2L μ
T 1 = 1.020T
Therefore f 1 = 1.020 f
= 1.020(440 Hz)
= 444 Hz

Beat frequency, f beat = f 2 – f 1
= (444 – 440) Hz
= 4.0 Hz


Section C Essay Question


Question 4
(a) Explain the meaning of beats. State the conditions for the beats to be heard.
(b) When the note C of a piano at 523 Hz is tuned using a reference oscillator,
–1
3.00 beats s are heard.
Semester
3 (i) What are the possible frequencies of the string?
(ii) When the string is tightens slightly, 4.00 beats s are heard. What is the
–1
frequency of the string?
416




021 STPM Q&A Physics T3.indd 416 17/02/2022 10:24 AM

Physics Semester 3 STPM Chapter 21 Sound Waves

(iii) By what percentage should the piano tuner change the tension in the
string to bring it into tune?
State whether the change is an increase or decrease.

Penerbitan Pelangi Sdn Bhd. All Rights Reserved.
Answer
(a) Beats are the periodic and repeating fluctuations in the intensity of a
sound when two sound waves of very similar frequencies interfere with
one another.
Conditions: Sound waves of slight difference in frequencies, and equal or
approximately equal amplitudes.

(b) (i) f 2 – 523 = 3.00 Hz, or 523 – f 2 = 3.00 Hz
f 2 = 526 Hz or f 2 = 520 Hz

(ii) f 2 = 1 T when T is increased, f 2 increased.
2L μ
Beat frequency increases to 4.00 Hz.
Therefore correct equation is f ' 2 – 523 = 4.00 Hz
Hence new frequency of string f ' 2 = 527 Hz

1 1 T
(iii) f ' 2 = T' 2 = 527 Hz , f = = 523 Hz
2L μ 2L μ
T 523 2
=  
T' 2 527
T = 0.985 T' 2
Tension in the string should be reduced by 1.5%.




21.5 Doppler Effect


Section A Multiple-choice Questions

Question 1
–1
A police car with its siren on travelled at a speed of 20 m s toward an observer.
The frequency of the siren is 900 Hz. What is the frequency heard by the Semester
–1
observer? (Speed of sound in air is 330 m s ).
A 845 Hz C 954 Hz
B 849 Hz D 958 Hz 3



417




021 STPM Q&A Physics T3.indd 417 17/02/2022 10:24 AM

Physics Semester 3 STPM Chapter 21 Sound Waves
Answer: D
Frequency increases if separation between observer and source decreases and
decreases when the separation increases.
Apparent frequency, f ’ =  v  f
v – u s
Penerbitan Pelangi Sdn Bhd. All Rights Reserved.
330
=  330 – 20 (900) Hz
= 958 Hz
Common Error
Do not confure the velocity u 0 of the observer, and u s the velocity of the source
in the equation apparent frequency, f ’ =  v ± u 0  f.
v  u s
If the observer and the source are getting closer, f’ > f, and if the separation
between them is increasng, f’ > f.

Question 2

An observer moves with a constant speed away from a stationary source that
emits sound of frequency f and wavelength λ. The sound detected by the
observer has frequency and wavelength
Frequency Wavelength
A less than f equals λ
B greater than f less than λ
C greater than f equals λ
D less than f greater than λ


Answer: A
When the observer moves away from the source, the frequency detected is
lower because the speed of sound relative to the observer decreases.
Question 3
A sound source of frequency 1.00 kHz moves at a speed of 30.0 m s toward
–1
a listener who moves at a speed of 20.0 m s in a direction away from the
–1
source. What is the apparent frequency heard by the listener? (Speed of sound
–1
= 330 m s )
A 857 Hz B 861 Hz C 1033 Hz D 1129 Hz

Answer: C
Frequency heard by the listener, f’ =  v – u o  f
Semester
v – u s
3 =  330 – 20.0  1000 Hz
330 – 30
= 1033 Hz

418




021 STPM Q&A Physics T3.indd 418 17/02/2022 10:24 AM

Physics Semester 3 STPM Chapter 21 Sound Waves
Question 4

A source of sound has a constant frequency. If all the motions described below
have the same speed, the observed frequency of sound is the highest in
A the source and observer are moving away from each other
Penerbitan Pelangi Sdn Bhd. All Rights Reserved.
B the source and observer are moving toward each other
C the source is moving toward a stationary observer
D the source is stationary and the observer is moving toward it


Answer: B
The separation between the source and observer decreases at the fastest rate
when they are moving toward each other.

Question 5

A submarine that emits sonar of frequency f is moving with a velocity u toward
a stationary object. The speed of sonar in water is v. The frequency of the sonar
reflected from the object and detected by the submarine is
A  v – u  f
v + u
B  v  f
v – u
C  v + u  f
v – u
D  v + u  f
v


Answer: C
Use the equation f’ =  v  u o  f
v  u s
v
Wave incident on stationary object, f 1 =  v – u  f
For the reflected wave, the submarine is the “observer”
v
f

frequency detected, f 2 =  v + u  1 =  v + u   v – u  f
v
v
=  v + u  f Semester
v – u
3




419




021 STPM Q&A Physics T3.indd 419 17/02/2022 10:24 AM

Physics Semester 3 STPM Chapter 21 Sound Waves

Section B Structured Questions

Question 6
An ambulance with its siren on is approaching a tall building with a speed of
Penerbitan Pelangi Sdn Bhd. All Rights Reserved.
25 m s . The ambulance is moving away from an observer who is in front of
–1
the building as shown in the figure. The frequency of the siren is 1200 Hz and
the speed of sound is 330 m s .
–1
Building

Ambulance Observer
25 m s –1
Calculate (a) the frequency of sound received by the observer,
(b) the frequency of sound reflected from the building, and
(c) the frequency of the beat detected by the observer.


Answer
(a) The ambulance (source) is moving away from the observer.
The frequency received by the observer is less than 1200 Hz.

f 1 =  v  f =  330  1200 Hz
v + u s 330 + 25
= 1115 Hz
(b) The source is approaching the wall. Frequency of sound incident on the
wall is higher than 1200 Hz. The reflected sound has the same frequency.

f 2 =  v  f =  330  1200 Hz
v – u s 330 – 25
= 1298 Hz

(c) Interference between the reflected sound and the sound from the
ambulance received by the observer produces beats of frequency
f = f 2 – f 1 = (1298 – 1115) Hz
= 183 Hz




Semester
3




420




021 STPM Q&A Physics T3.indd 420 17/02/2022 10:24 AM

Physics Semester 3 STPM Chapter 21 Sound Waves

Section C Essay Question
Question 7

(a) What is meant by Doppler Effect?
Penerbitan Pelangi Sdn Bhd. All Rights Reserved.
(b) A sound source emits sound of constant frequency. Draw three circles
to represent three successive wavefronts from the source in each of the
following situations.
(i) The source is stationary.
(ii) The source moves toward an observer with a speed less than the speed
of sound.
(iii) The source moves toward an observer with a speed that equals to the
speed of sound.
(c) From the diagrams in (b)(ii), deduce the changes in the wavelength and
frequency detected by the observer.
(d) Explain why a loud sound is heard by the observer in (b)(iii).
(e) An ambulance moving with a velocity of 42 m s sounds its siren of
–1
frequency 500 Hz. A car is moving in the same direction as the ambulance
–1
–1
at 25 m s . The speed of the sound is 330 m s .
What frequency does a person in the car hear
(i) as the ambulance approaches the car?
(ii) after the ambulance passes the car?

Answer
(a) Doppler effect: The change in the frequency of sound heard by an
observer when there is relative motion between the sound source and the
observer.

(b)

0
<
u
S S u < v v S
u=v
(i) (ii) (iii)

(c) The wavelength is shorter. The frequency increases.
(d) The wavefronts overlap. Constructive superposition that results in the
sound wave received by the observer has a high amplitude. Semester



3




421




021 STPM Q&A Physics T3.indd 421 17/02/2022 10:24 AM

Physics Semester 3 STPM Chapter 21 Sound Waves
(e) (i) The ambulance (source) is approaching the car (observer), and the
car is moving away from the ambulance.
u = 42.0 m s –1 u = 25.0 m s –1
s o

Penerbitan Pelangi Sdn Bhd. All Rights Reserved.


Frequency heard, f 1 =  330 – 25.0  (500 Hz) = 530 Hz
330 – 42.0

(ii) The ambulance (source) is now moving away from the car (observer),
and the car is approaching the ambulance.

u = 25.0 m s –1 u = 42.0 m s –1
s
0




Frequency heard, f 1 =  330 + 25.0  (500 Hz) = 477 Hz
330 + 42.0































Semester
3




422




021 STPM Q&A Physics T3.indd 422 17/02/2022 10:24 AM

ANSWERS




STPM Model Paper 960/1
Penerbitan Pelangi Sdn Bhd. All Rights Reserved.
Section A
1. D: Average time = 24.475 s
t = (24.5 ± 0.1) s
20
2. B: u = 0 + at . v = 0 + at 2
1
v – u
a =
t – t 1
2
Semester
T
1
1 3. A: H = (u + 0)( ) = uT
2 2 4
4. C: (v cos 60°)(1.0) = 6.0 m, gives v = 12.0 m s –1
1
y = (12.0 sin 60°)(1.0) – g(1.0) 2
1 2
= 5.5 m
5. C: F = 12.0 N = (1.0 + 2.0 + 3.0)a
T = (1.0)(2.0) N = 2.0 N
2
T – T = (2.0)(2.0) N, gives T = 6.0 N
1
2
1
6. A: At 7.5 m s , F = (1.5 )(600 N)
2
–1
Power, P = Fv = 10.1 kW
7. B: Geo-synchronous orbit is above the equator and at the same direction as the
rotation of the Earth about its axis.
1 GMm GMm
2
8. C: mu = (– ) – (– )
2 3R R
u = 4GM
3R
9. D: Moments about the centre of the square = 0
Resultant force = 0
10. A
EA
11. A: F = kx = ( l )x
k = 2k
P Q
E A E A
P = 2( Q )
2l l
E
P = 4.0
E
Q
1 1 3
12. B: M c = M c = RT
2
2
2 H H 2 He He 2
c M
H = He = 1.4
c
He M
H
536
ANS STPM Q&A Physics.indd 536 16/02/2022 4:35 PM

Physics STPM Answers
13. D: XY:Q = ∆U + W, Q = 0
∆U = –W = –15 J
At Y, U = (50 – 15) J = 35 J

Temperature at X and Z is equal. U = U = 50 J
x z
Penerbitan Pelangi Sdn Bhd. All Rights Reserved.
YZ: Q = ∆U + W, W = 0
= U – U + 0
Z Y
= (50 – 35) J = +15 J
14. D: Stefan’s law: P = eσAT 4 Semester
k (q – 40)/2l
15. B: = gives
2k (100 – q)/l 1
Temperature at junction, q = 70°C

Section B
16. (a) (2.5 + 1.5)v = (2.5)(0.50) – (1.5)(0.30)
v = 0.20 m s
–1
(b) (2.5)v + (1.5)v = (2.5)(0.50) – (1.5)(0.30)
2
1
1
1
1
1 (2.5)v + (1.5)v = (2.5)(0.50) + (1.5)(0.30) 2
2
2
2
2 1 2 2 2 2

Alternative method:
(2.5)v + (1.5)v = (2.5)(0.50) – (1.5)(0.30)
1 2
Newton’s law of restitution:
v – v = –[0.50 – (–0.30)]
2
1
Solving, = –0.10 m s , v = 0.70 m s –1
v
–1
1 2
E A E A
17. (a) F = ( Cu l )e = ( Fe )e Fe
Cu
2l
e
e Cu = 1.0
Fe
(b) F = k e = k e
Cu
Cu
Fe Fe
k
Cu = 1.0
k Fe
e F
(c) (Strain) = Cu =
Cu
l E A
F
e Cu
(Strain) = Fe =
Fe 2l E A
Fe
(Strain) / (Strain) = 2
Cu Fe
1
(d) Strain energy = Fe
2
(Strain energy) = (Strain energy)
Cu Fe

537
ANS STPM Q&A Physics.indd 537 16/02/2022 4:35 PM

Physics STPM Answers
Section C
18. (a) Centripetal force: force towards the centre of the circle that enables the particle
to move in a circle.
Reason for the acceleration:
• direction of velocity is constantly changing
Penerbitan Pelangi Sdn Bhd. All Rights Reserved.
• acceleration = rates of change of velocity
(b) (i)
Friction, F  Reaction, R



Semester
1 Weight, mg
(ii) F = μR
R sin 30° + (μR) cos 30° = mg
(0.050)(9.81)
R = N = 0.481 N
sin 30° + (0.60)cos 30°
(iii) Friction = μR = 0.289 N
(iv) R cos 30° – F sin 30° = (0.050)r[2(2.5)π] 2
(0.481)cos 30° – (0.289)sin 30°
Radius, r = m
(0.050)(5.0π) 2
= 2.20 cm
(v) Centripetal force, R cos 30° – F sin 30° = constant
= mrw 2
When r decreases, w increases.
19. (a) (i) n : number of gas molecules per unit volume
m : mass of gas molecule
<c >: mean square speed of gas molecules
2
1 1 1 M
(ii) p = nm<c > = ρ<c > = <c >
2
2
2
3 3 3 V m
1
pV = (mN )<c >= RT
2
m 3 A
3
3 R
1
Mean translational K.E. = m<c > = ( )T = kT
2
2 N
2
2
A
(b) (i) p (iii) I: W= p (2V – V ) = p V
o 0 0 o 0
I 2V
II: W = nRT ln ( 0 ),
0
II V 0
p V = nRT 0
o
0
III
0 V W = p V ln2
o
0
V 0 2V 0
538
ANS STPM Q&A Physics.indd 538 16/02/2022 4:35 PM

Format: 146mm X 216mm Extent= 560 pgs (26.7mm) 70gsm Status: CRC Date: 22/2
Q A EC059242 PELANGI BESTSELLER


&




STPM
A
Penerbitan Pelangi Sdn Bhd. All Rights Reserved.

. .
Semester 1 2 3 Q & A STPM Q

Physics [960] &



Q & A STPM fulfils the needs of students in mastering the
technique of answering questions effectively to excel in the
STPM exam. The questions and answers provided meet the
STPM exam standard and are useful as a revision tool.



FEATURES TITLES IN THIS SERIES: Physics
• Based on the latest syllabus ■ Pengajian Am
• Arranged according to subtopics ■ Bahasa Melayu
• Tips to guide students in answering ■ Sejarah
questions effectively ■ Geografi
• Common Errors to show common ■ Ekonomi
mistakes frequently made by students ■ Pengajian Perniagaan STPM
while answering questions ■ Mathematics (T)
• Model Paper which follows the latest ■ Physics
STPM assessment format to prepare ■ Chemistry
students for the actual exam ■ Biology . .
Pengajian Am Ekonomi Physics Semester 1 . 2 . 3 Semester 1 2 3
Bahasa Melayu Pengajian Perniagaan Chemistry
Sejarah Mathematics (T) Biology
Physics
Geograf


Purchase
[960]
eBook here!

W.M: RM31.95 / E.M: RM32.95
EC059242
ISBN: 978-967-0007-34-2



PELANGI Lin Poh Tin • Poh Liong Yong